//<p><strong>路径</strong> 被定义为一条从树中任意节点出发，沿父节点-子节点连接，达到任意节点的序列。同一个节点在一条路径序列中 <strong>至多出现一次</strong> 。该路径<strong> 至少包含一个 </strong>节点，且不一定经过根节点。</p>
//
//<p><strong>路径和</strong> 是路径中各节点值的总和。</p>
//
//<p>给你一个二叉树的根节点 <code>root</code> ，返回其 <strong>最大路径和</strong> 。</p>
//
//<p> </p>
//
//<p><strong>示例 1：</strong></p>
//<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg" style="width: 322px; height: 182px;" />
//<pre>
//<strong>输入：</strong>root = [1,2,3]
//<strong>输出：</strong>6
//<strong>解释：</strong>最优路径是 2 -> 1 -> 3 ，路径和为 2 + 1 + 3 = 6</pre>
//
//<p><strong>示例 2：</strong></p>
//<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg" />
//<pre>
//<strong>输入：</strong>root = [-10,9,20,null,null,15,7]
//<strong>输出：</strong>42
//<strong>解释：</strong>最优路径是 15 -> 20 -> 7 ，路径和为 15 + 20 + 7 = 42
//</pre>
//
//<p> </p>
//
//<p><strong>提示：</strong></p>
//
//<ul>
//	<li>树中节点数目范围是 <code>[1, 3 * 10<sup>4</sup>]</code></li>
//	<li><code>-1000 <= Node.val <= 1000</code></li>
//</ul>
//<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>动态规划</li><li>二叉树</li></div></div><br><div><li>👍 1631</li><li>👎 0</li></div>

package com.rising.leetcode.editor.cn;

import com.rising.leetcode.editor.cn.doc.object.TreeNode;

/**
 * 二叉树中的最大路径和
 * @author DY Rising
 * @date 2022-06-23 20:47:16
 */
public class P124_BinaryTreeMaximumPathSum{
    public static void main(String[] args) {
        //测试代码
        Solution solution = new P124_BinaryTreeMaximumPathSum().new Solution();
    }
	 
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int maxSum = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        maxSum(root);
        return maxSum;
    }

    /**
     * 计算以某一节点为根节点的最大贡献值
     * @param node
     * @return
     */
    public int maxSum(TreeNode node) {
        if (node == null) {
            return 0;
        }

        int leftMaxSum = Math.max(maxSum(node.left), 0);
        int rightMaxSum = Math.max(maxSum(node.right), 0);

        int maxPathSum = node.val + leftMaxSum + rightMaxSum;

        maxSum = Math.max(maxPathSum, maxSum);

        return node.val + Math.max(leftMaxSum, rightMaxSum);
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}
